How to Solve a Mid-Problem Calculus Step?

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Discussion Overview

The discussion focuses on a specific step in a calculus problem involving the differentiation of a function y with respect to x, particularly the expression \(\frac {d}{dx} ( \frac {y} {\sqrt{1 + y'}} )\). Participants explore the appropriate differentiation techniques and clarify notation used in calculus.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on whether to use the quotient rule for differentiation and questions if \(\frac {d}{dx}y\) equals \(y'\).
  • Another participant confirms that \(\frac{dy}{dx}\) is equivalent to \(y'\) and explains the difference between Leibniz's and Lagrange's notation.
  • A third participant suggests using the quotient rule and mentions the possibility of applying the chain rule for the square root in the denominator.
  • A later reply emphasizes that since the expression involves a function of y, each derivative must be multiplied by \(y'\) when differentiating with respect to x.

Areas of Agreement / Disagreement

Participants generally agree on the use of the quotient rule and the equivalence of different notations for derivatives, but there is no consensus on the specific steps or methods to apply in this context.

Contextual Notes

Participants do not explicitly address any missing assumptions or unresolved mathematical steps in their discussion.

ElDavidas
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Hi,

could someone please clarrify how to do this step? It's midway through a large problem and I'm not so sure about it.

y is a function of x.

[tex]\frac {d}{dx} ( \frac {y} \sqrt{1 + y'}} )[/tex]

Please excuse the Latex. The square root of 1 + y' is being taken on the denominator .

Do you just use the quotient rule? and does [itex]\frac {d} {dx}y[/itex] equal [itex]y'[/itex]?

thanks
 
Last edited:
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I can't see the LaTeX but dy/dx does equal y'.

dy/dx is Leibniz's notation and using the prime mark is Lagrange's notation.

f(x)=y, so f'(x)=y' and if y is a function of x then the dy/dx means the derivative of y with respect to x.
 
Last edited:
Yes, use the quotient rule. To differentiate the square root expression you could use the chain rule. It may help to look at the denominator as [tex](1+y')^\frac{1}{2}[/tex].
 
Further, since that is a function of y and you are differentiating with respect to x, you will have to multiply each derivative by y'.
 

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